Question

Let $\phi(x) = 3f(\frac{x^2}{3})+f(3-x^2)\forall x \in (-3,4)$ where $f''(x)>0 \forall x \in (-3,4), g(x) = (a^2-3a+2)(\cos^2\frac{x}{4}-\sin^2\frac{x}{4})+(a-1)x+\sin 1$ Match List-I with List-II and select the correct answer using the code given below the list.

• A. $\phi(x)$ is increasing in
• B. $\phi(x)$ is decreasing in
• C. $(0, 1) \cup (1, 4)$
• D. $[-\frac{3}{2}, 0];[\frac{3}{2}, 4)$

Answer 1

Answer: (D)

P matches with 2

Answer 2

To determine the intervals where $\phi(x)$ is increasing or decreasing, we compute its derivative $\phi'(x)$.

Given $\phi(x) = 3f(\frac{x^2}{3}) + f(3-x^2)$, we find: $\phi'(x) = 2x [ f'(\frac{x^2}{3}) - f'(3-x^2) ]$

Since $f''(x) > 0$ for all $x \in (-3, 4)$, $f'(x)$ is strictly increasing on $(-3, 4)$.

The sign of $f'(\frac{x^2}{3}) - f'(3-x^2)$ is the same as the sign of $\frac{x^2}{3} - (3-x^2) = \frac{4x^2}{3} - 3$.

So, the sign of $\phi'(x)$ is the same as the sign of $2x(\frac{4x^2}{3} - 3)$.

We find the critical points by setting $\phi'(x) = 0$, which gives us $x = 0$ and $x = \pm \frac{3}{2}$.

Analyzing the sign of $\phi'(x)$ in the intervals defined by these critical points:

• For $x \in (-\frac{3}{2}, 0)$, $\phi'(x) > 0$, so $\phi(x)$ is increasing.

• For $x \in (\frac{3}{2}, 4)$, $\phi'(x) > 0$, so $\phi(x)$ is increasing.

Therefore, $\phi(x)$ is increasing on $(-\frac{3}{2}, 0) \cup (\frac{3}{2}, 4)$. Including endpoints where the derivative is zero, it is increasing on $[-\frac{3}{2}, 0]$ and $[\frac{3}{2}, 4)$. This matches option (2).